CBSE Class 9

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC

Solution:
(i) In ∆ABD and ∆ACD
AB = AC (∵ ABC is an isosceles ∆)
BD = CD (∵ DBC is an isosceles ∆)
AD = AD (Common)
By S-S-S Congruency condition
∆ABD = ∆ACD
∴ ∠BAD = ∠CAD ……(A) (By C.P.CT)

(ii) In ∆ABP and ∆ACP
AB = AC (∵ ABC is an isoscles ∆)
∠BAP = ∠CAP
AP = AP
By S-A-S Congruency Condition
∠ABP = ∠ACP
Therefore, BP = CP …….(i) (By C.P.C.T.)

(iii) We have
∆ABD = ∆ACD (From (A))
∴ ∠BAD = ∠CAD (By C.P.C.T.)
∴ AP bisects ∠A
Again,
In ∆BDP and ∆GDP
BD = CD (∵ ∆BDC is an isoscles ∆)
DP = DP (Common)
BP = CP (From equation (i))
By S-S-S Congruency Condition
∆BDP ≅ ∆CDP ………(B)
∴ ∠BDP = ∠CDP (By C.P.C.T)
Therefore, AP bisects ∠D …….(iii)
From equation (ii) and (iii) we can say AP bisects ∠A as well as ∠D.

(iv) we have
∆BDP = ∆CDP (from (B))
∴ BP = CP ……(iv) (By C.P.C.T.)
Again, ∠BPD = ∠CPD (By C.P.C.T.)
Now, ∠BPD + ∠CPD = 180
or, ∠BPD + ∠BPD = 180
2∠BPD = 180
∴ ∠BPD = 90 ……(v)
Therefore, From equations (iv) and (v) we can say AP is the perpendicular bisector of BC.

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A

Solution:
(i) In ∆ABD and ∆ACD
AB = AC (Given)
∠ADB = ∠ADC (each 90°)
AD = AD (Common)
By R.H.S Congruency Condition
∆ABD = ∆ACD ……(i)
∴ BD = CD (By C.P.C.T.)
Therefore, AD bisects BC.

(ii) We have
∆ABD = ∆ACD (from equation (i))
∴ ∠BAD = ∠CAD (By C.P.C.T.)
Therefore, AD bisects ∠A.

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see Fig. 7.40). Show that:
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR

Solution:
(i) In ∆ABM and ∆PQN
AB = PQ (Given)
BM = QN (Given)
AM = PN (Given)
By S-S-S Congruency Condition,
∆ABM ≅ ∆PQN …..prove (i)
∴ ∠B = ∠Q (By C.P.C.T)

(ii) Now, In ∆ABC and ∆PQR
AB = PQ (Given)
BM = QN (Given)
∴ 2BM = 2QN
or, BC = QR
(Because M and N are respectively the midpoint of side BC and QR)
∠B = ∠Q (Prove above)
By S-A-S Congruency Condition
∆ABC ≅ ∆PQR …….prove (ii)

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using the R.H.S congruence rule, prove that the triangle ABC is isosceles.

Solution:
In ∆BFC and ∆CEB
CR = BE (Given)
∠BFC = ∠CEB (Each 90°)
BC = BC (Common)
∴ By R-H-S congruency condition,
∆BFC ≅ ∆CEB
∴ ∠B = ∠C (By C.P.C.T)
Therefore, AB = AC (Sides opposite to equal angles of a ∆)
Hence, ∆ABC is isosceles.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution:
In ∆ABP and ∆ACP
AB = AC (Given)
∠APB = ∠APC (Each 90°)
AP = AP (Common)
By R-H-S Congruency Condition,
∆ABP ≅ ∆ACP
or, ∠B = ∠C (By C.P.C.T)

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2

Question 1.
In an issosceies triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O show that:
(i) OB = OC
(ii) AO bisects ∠A
Solution:
(i) In ∆ABC,
We have given that
AB = AC
∴ ∠B = ∠C (Angle opposite to equal sides are equal)
or, \(\frac {1}{2}\) ∠B = \(\frac {1}{2}\) ∠C
∴ ∠1 = ∠2
Now in ∠OBC,
∠1 = ∠2
∴ OB = OC (Side opposite to equal angles are equal)

(ii) In ∆AOB and ∆AOC
AB = AC
OB = OC
∠B = ∠C
∴ \(\frac {1}{2}\) ∠B = \(\frac {1}{2}\) ∠B
or ∠3 = ∠4
Therefore, by S-A-S Congruency Condition,
∆AOB ≅ ∆AOC
∴ ∠BAO = ∠CAO (by C.P.C.T)
∴ AO is the bisector of ∠A.

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ∆ABC is an isosceles triangle in which AB = AC.

Solution:
In ∆ABD and ∆ACD
BD = CD (∵ AD bisects BC)
∠ADB = ∠ADC (Each 90°)
AD = AD (Common)
By S-A-S Congurency Condition,
∆ABD ≅ ∆ACD
Therefore, AB = AC (By C.P.C.T)
∴ ABC is an isosceles triangle.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to side AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Solution:
In ∆ABE and ∆ACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (each 90°)
AB = AC (Given)
By, A-A-S Congruency Condition
∆ABE = ∆ACP
Therefore, BE = CF (By C.P.C.T)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC, i.e. ∆ABC is an isosceles triangle.

Solution:
(i) ∆ABE and ∆ACF,
BE = CF (Given)
∠A = ∠A (Common)
∠AEB = ∠AFC (Each 90°)
By A-A-S Congruency Condition
∆ABE = ∆ACF

(ii) Since ∆ABE ≅ ∆ACF
So, AB = AC (By C.P.C.T.)
or, ∆ABC is an isosceles triangle.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

Solution:
In ∆ABC,
AB = AC
∴ ∠ABC = ∠ACB …….(i)
(Angle opposite to equal sides of a ∆ are equal)
Again, In ∆DBC
DB = DC
∠DBC = ∠DCB ……(ii)
(Angle opposite to equal sides of a ∆ are equal)
Adding equation (i) and (ii)
∠ABC + ∠DBC = ∠ACB + ∠DCB
or ∠ABD = ∠ACD.

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

Solution:
In ∆ABC,
AB = AC
∠ABC = ∠ACB ……(i)
(Angles opposite to equal sides of a ∆ABC)
Now, In ∆ACD,
AC = AD
∠ACD = ∠ADC ……(ii)
(Angles opposite to equal sides of ∆ACD)
Now, ∠BAC + ∠CAD = 180° ……(iii) (Linear pair)
Also, ∠CAD = ∠ABC + ∠ACB (Exterior angle of ∆ABC)
∠CAD = 2∠ACB …….(iv)
From equation (i),
∠ABC = ∠ACB
and ∠BAC = ∠ACD + ∠ADC (Exterior angle of ∆ADC)
∠BAC = 2∠ACD …….(v)
(From equation (ii), ∠ACD = ∠ADC)
From equation (iii) we have,
∠BAC + ∠CAD = 180°
⇒ 2∠ACD + 2∠ACB = 180° (From equation (iv) and (v))
⇒ 2(∠ACD + ∠ACB) = 180°
⇒ 2(∠BCD) = 180°
⇒ ∠BCD = 90°

Question 7.
ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Solution:
In ∆ABC,
AB = AC
∴ ∠B = ∠C (Angle opposite to equal sides of ∆ABC)
Now, ∠A + ∠B + ∠C = 180° (Angle sum property)
or, 90° + ∠B + ∠B = 180° (∵ ∠A = 90° and ∠B = ∠C)
or, 2∠B = 90°
∴ ∠B = 45°
Therefore, ∠B = ∠C = 45°.

Question 8.
Show that the angles of equilateral triangles are 60° each.

Solution:
In ∆ABC,
AB = BC = AC (Because ABC is an equilateral triangle)
∴ ∠C = ∠A = ∠B (Angle opposite to equal sides of ∆ABC)
Now, ∠A + ∠B + ∠C = 180° (Angle sum property of ∆)
or, ∠A + ∠A + ∠A = 180° (∵ ∠A = ∠B = ∠C)
or, 3∠A = 180
Therefore, ∠A = 60°
Again, ∠A = ∠B = ∠C
∴ ∠A = ∠B = ∠C = 60°
Therefore, each angle of an equilateral ∆ is 60°.

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1

Question 1.
In quadrilateral ACBD (See Fig. 7.16), AC = AD, and AB bisects ∠A. Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

Solution:
In ΔABC and ΔABD
AC = AD (Given)
∠CAB = ∠DAB (Given)
AB = AB (Common)
Therefore, By SAS congruency condition
ΔABC ≅ ΔABD
So, BC = BD (By C.P.C.T)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17)
Prove that:
(i) ΔABD = ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC

Solution:
(i) In ΔABD and ΔBAC
AD = BC
∠DAB = ∠CBA (Given)
and AB = BA (Common)
By SAS Congruency Condition
ΔABD = ΔBAC
(ii) BD= AC (By C.P.C.T)
(iii) ∠ABD = ∠BAC (Again by C.P.C.T)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Solution:
In ΔAOD and ΔBOC,
AD = BC (Given)
∠OAD = ∠OBC (each 90°)
∠AOD = ∠BOC (Vertically opposite angles)
Therefore, by ASA congruency condition.
ΔAOD = ΔBOC
So, OA = OB (by C.P.C.T)
Hence, CD bisects line segment AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC = ΔCDA.

Solution:
We have given that l || m and p || q
Therefore, In ΔABC and ΔCDA
∠BAC = ∠DCA
(Alternate interior angles as AB || DC)
∠ACB = ∠CAD
(Alternate interior angles as BC || DA)
AC = CA (common)
So, By A-S-A congruency condition,
ΔABC = ΔCDA

Question 5.
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20) show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistance from the arms of ∠A.

Solution:
In ΔABP and ΔABQ,
∠BAP = ∠BAQ (Given)
∠APB = ∠AQB (Each 90°)
AB = AB (Common)
By A-A-S congruency condition.
So, ΔABP ≅ ΔABQ
(ii) BP = BQ (By C.P.C.T)

Question 6.
In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
In ΔBAC and ΔDAE
AB = AD (Given)
AC = AE (Given)
∠BAD = ∠EAC ……(i) (Given)
Adding ∠DAC both side in equation (i)
∴ ∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
Therefore by S-A-S Congruency Condition
ΔBAC = ΔDAE
So, BO = DE (By C.P.C.T)

Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22) show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

Solution:
(i) In ΔDAP and ΔEBP
∠DAP= ∠EBP (Given)
∠APE = ∠DPB (Given)
∴ ∠APE + ∠EPD = ∠DPB + ∠EPD (Add ∠EPD both side)
∠APD = ∠BPE
AP = BP (Given P is the mid point of AB)
∴ By A-S-A Congruency Condition,
ΔDAP ≅ ΔEBP
(ii) AD = BE (By C.P.C.T.)

Question 8.
In the right triangle ABC right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produces to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = \(\frac {1}{2}\) AB

Solution:
(i) In ΔAMC and ΔBMD,
AM = BM (Given)
CM = DM (Given)
∠AMC = ∠BMD (Vertically opposite angles)
∴ By S-A-S Congruency Condition.
ΔAMC = ΔBMD

(ii) ∠CAM = ∠DBM (by C.P.C.T)
Also, ∠CAM + ∠MBC = 90° (Since ∠C = 90°)
∴ ∠DBM + ∠MBC = 90° (∵ ∠CAM = ∠DBM)
or, ∠DBC = 90°

(iii) In ΔDBC and ΔACB,
BC = BC (Common)
DB = AC (∵ ΔBMD ≅ ΔAMC by C.PC.T)
and ∠DBC = ∠ACB (each 90° proved above)
Therefore, by S-A-S Congruency Condition,
∴ ΔDBC ≅ ΔACB

(iv) Since, ΔDBC ≅ ΔACB,
DC = AB
∴ \(\frac {1}{2}\) DC = \(\frac {1}{2}\) AB
CM = AM
or CM = \(\frac {1}{2}\) AB

These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.3

Question 1.
In Fig. 6.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135 and ∠PQT = 110°, Find ∠PRQ.

Solution:
In ∆PQR
∠SPR + ∠RPQ = 180° (Linear pair)
or, ∠RPQ = 180° – 135° (∵ ∠SPR = 135°)
or, ∠RPQ = 45° …….(i)
Again, ∠PQT + ∠PQR = 180° (Linear pair)
or, 110° + ∠PQR = 180° (∵ ∠PQT = 110°)
∠PQR = 180° – 110°
or, ∠PQR = 70° ……(ii)
Now, we know that sum of all angles of a ∆ is 180°
Therefore,
∠RPQ + ∠PQR + ∠PRQ = 180°
or, 45° + 70° + ∠RPQ = 180° [From equ. (i) and (ii)]
∠RPQ = 180° – 45° – 70° = 65°
Therefore, ∠RPQ = 65°

Question 2.
In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisector of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.

Solution:
In ∆XYZ,
∠X + ∠XYZ + ∠XZY = 180° (Angle sum property of ∆)
or, 62° + 54° + XZY = 180° (∴ ∠X = 65° & ∠XYZ = 54°)
or, ∠XYZ = 180° – 116°
or, ∠XZY = 64° ….(i)
Again, ∠XZY = ∠XZO + ∠OZY (∴ OZ is the bisector of ∠XZY)
or, 2∠OZY = 64°
or, ∠OZY = 32°
∠OYZ + ∠OZY + ∠YOZ = 180° (Angle sum property of ∆)
27° + 32° + ∠YOZ = 180°
or, ∠YOZ = 121°

Question 3.
In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:
We have given AB || DE
∠BAC = ∠DEC (Pair of alternate interior angles)
or, ∠DEC = 35° (∵ ∠BAC = 35°)
Now, In ∆CDE
∠CDE + ∠CED + ∠DCE = 180° (Angle sum property of ∆)
⇒ 53 + 35 + ∠DCE = 180°
⇒ ∠DCE = 180° – 53° – 35°
⇒ ∠DCE = 92°

Question 4.
In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:
In ∆PRT,
∠RPT + ∠PRT + ∠PTR = 180° (Angle sum property of ∆)
95 + 40 + ∠PTR = 180°
or, ∠PTR = 180° – 95° – 40°
or, ∠PTR = 45°
Again, ∠PTR = ∠QTS (Vertically opposite angles)
∴ ∠QTS = 45°
Now, In ∆QTS,
∠QTS + ∠TSQ + ∠SQT = 180° (Angle sum property of ∆)
45° + 75° + ∠SQT = 180° (∵ ∠QTS = 45° and ∠TSQ = 75°)
or, ∠SQT = 180° – 45° – 75°
∴ ∠SQT = 60°

Question 5.
In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the value of x and y.

Solution:
We have given PQ || SR
∴ ∠PQR = ∠QRT (Pair of alternate interior angles)
⇒ x + 28° = 65°
⇒ x = 65° – 28°
⇒ x = 37°
Now, in ∆PQS
∠QPS + x + y = 180° (Angle sum property of ∆)
⇒ 90° + 37° + y = 180° (∴ PQ ⊥ PS and x = 37)
⇒ y = 180° – 90° – 37°
⇒ y = 53°

Question 6.
In Fig. 6.44, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac {1}{2}\) ∠QPR.

Solution:
In ∆PQR, side QR is produced to S, so by exterior angle property,
∠PRS = ∠P + ∠PQR
⇒ \(\frac {1}{2}\) ∠PRS = \(\frac {1}{2}\) ∠P + \(\frac {1}{2}\) ∠PQR
⇒ ∠TRS = \(\frac {1}{2}\) ∠P + ∠TQR ……(1)
[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]
Now, in ∆QRT, we have
∠TRS = ∠TQR + ∠T ……(2)
[Exterior angle property of a triangle]
From (1) and (2),
we have ∠TQR + \(\frac {1}{2}\) ∠P = ∠TQR + ∠T
⇒ \(\frac {1}{2}\) ∠P = ∠T
⇒ \(\frac {1}{2}\) ∠QPR = ∠QTR or ∠QTR = \(\frac {1}{2}\) ∠QPR

These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.2

Question 1.
In Fig. 6.28, find the values of x and y and then show that AB || CD.

Solution:
According to fig.
50° + x = 180° (Linear pair)
⇒ x = 180° – 50°
⇒ x = 130° …..(i)
Again, x = 130° …..(ii)
(Vertically opposite angles are equal)
From equation (i) and (ii)
x = y (each 130°)
which is the pair of alternate interior angles. And we know that if the pair of alternate interior angles are equal, then the given two lines are parallel.
AB || CD

Question 2.
In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:
We have given AB || CD || EF and y : z = 3 : 7
or, \(\frac{y}{z}=\frac{3}{7}\)
or, y = \(\frac {3}{7}\) z …….(i)
Again AB || EF
∴ x = z ……(ii)
(Pair of alternate interior angle)
Now, AB || CD
∴ x + y = 180°
(Sum of interior angle of the same side of transversal)
z + \(\frac {3}{7}\) z = 180 (∴ x = z and y = \(\frac {3}{7}\) z)
⇒ z = \(\frac{180 \times 7}{10}\)
⇒ z = 126
From, equation (ii)
x = z
or, x = 126

Question 3.
In Fig. 6.30, if AB || CD, EF || CD and ∠GED = 126°, Find ∠AGE, ∠CEF and ∠FGE.

Solution:
We have given AB || CD, EF || CD and ∠GED = 126°
Again ∠GED = ∠AGE (Alternate interior angle)
∠AGE = 126° (i) (∴ Given ∠GED = 126°)
Now, ∠GED = ∠DEF + ∠GEF
⇒ 126 = 90 + ∠GEF (∴ ∠GED = 126 & EF ⊥ CD)
⇒ ∠GEF = 126° – 90° = 36°
Again, ∠AGE + ∠FGE = 180° (Linear pair)
126° + ∠FGE = 180° (From equ. (i) ∠AGE = 126°)
∠FGE = 180° – 126° = 54°

Question 4.
In Fig. 6.31, if PQ || ST, ∠PQR = 110°, and ∠RST = 130°, find ∠QRS.

Solution:
We have given that PQ || ST,
∠PQR = 110° and ∠RST = 130°
Construction:
Through R draw a line MN | ST

Now, ST || RN (by construction)
Figure
Therefore,
∠RST + ∠SRN = 180° (Sum of interior angle of the same side of transversal)
or, 130° + ∠SRN = 180°
∠SRN = 180° – 130° = 50° ……(i)
Given ∠PQR = 110°
∠QRN = 110°
Again, ∠QRN = ∠QRS + ∠SRN
110 = ∠QRS + 50 (∵ ∠QRN = 110° and ∠SRN = 50° from equ. (i))
∴ ∠QRS = 110° – 50° = 60°

Question 5.
In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, Find x and y.

Solution:
We have given AB || CD
and ∠APQ = 50° & ∠PRD = 127°
Now, AB || CD (given)
∴ ∠APQ = x (Pair of alternate interior angle)
∴ x = 50° (∴ ∠APQ = 50°)
Again, ∠APR = ∠PRD (Pair of alternate interior angles)
∴ ∠APQ = 127°
But, ∠APR = 50° + y
⇒ 127° = 50° + y
⇒ y = 127° – 50°
⇒ y = 77°

Question 6.
In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Solution:
Two plane mirrors PQ and RS such that PQ || RS. An incident ray AB after reflections takes the path BC and CD. BM and CN are the normals to the plane mirror PQ and RS respectively.

To prove: AB || CD
Proof: Since BM ⊥ PQ, CN ⊥ RS, and PQ || RS
Therefore, CN ⊥ PQ ⇒ BM || CN
Thus, BM and CN are two parallel lines and a transversal BC cuts them at B and C respectively.
∴ ∠2 = ∠3 (Alternate interior angles)
But, ∠1 = ∠2 and ∠3 = ∠4 (By law of reflection)
⇒ ∠1 + ∠2 = ∠2 + ∠2 and ∠3 + ∠4 = ∠3 + ∠3
⇒ ∠1 + ∠2 = 2(∠2) and ∠3 + ∠4 = 2(∠3)
⇒ ∠1 + ∠2 = ∠3 + ∠4 [∵ ∠2 + ∠3 ⇒ 2(∠2) = (∠3)]
∴ ∠ABC = ∠BCD
Thus, lines AB and CD are intersected by transversal BC such that ∠ABC = ∠BCD i.e. alternate interior angles are equal.
Therefore, AB || CD
Hence, AB || CD.

These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1

Question 1.
In Fig. 6.13 lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°. Find ∠BOE and reflex ∠COE.

Solution:
In this fig. we have given that
∠AOC + ∠BOE = 70° and ∠BOD = 40°
But, ∠BOD = ∠AOC (Vertical opposite angles)
∠AOC = 40 ……(i) (∵ ∠BOD = 40°)
Again, ∠AOC + ∠BOE = 70° (Given)
40 + ∠ZBOE = 70° (from equation (i))
∠BOE = 70 – 40 = 30°
Again, ∠AOC + ∠COE + ∠EOB = 180° (Because AB is a st. line)
40° + ∠COE + 30° = 180° (∵ ∠AOC = 40° and ∠EOB = 30°)
Therefore, ∠BOE = 30°, and Reflex ∠COE = 360° – 110° = 250°
Therefore, ∠BOE = 30°, and Reflex ∠COE = 250°.

Question 2.
In Fig. 6.14 lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Solution:
We have given
∠POY = 90°, and a : b = 2 : 3
or, \(\frac{a}{b}=\frac{2}{3}\)
∴ a = \(\frac {2}{3}\) b ……(i)
Again, ∠POY + a + b = 180° (Because OXY is a straight line)
90 + \(\frac {2}{3}\) b + b = 180°
or, \(\frac {2}{3}\) b + b = 90
b = \(\frac{90 \times 3}{5}\) = 54° ……(ii)
Again, c + b = 180° (because MN is a straight line)
c + 54° = 180° (from equation (ii))
Therefore, c = 126°

Question 3.
In Fig. 6.15, ∠PQR = ∠PRQ then prove that ∠PQS = ∠PRT.

Solution:
We have given
∠PQR = ∠PRQ
Again,
∠PQR + ∠PQS = 180° ……(i) (Linear pair)
∠PRQ = ∠PRT = 180° ……(ii) (Linear pair)
From equation (i) and (ii)
∠PQR + ∠PQS = ∠PRQ + ∠PRT
But, ∠PQR = ∠PRQ
Therefore, ∠PQS = ∠PRT

Question 4.
In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Solution:
We have given x + y = w + z
Again, x + y + w + z = 360
or, x + y + x + y = 360 (∵ x + y = w + z)
or, 2(x + y) = 360°
or, x + y = 180°
∠AOB = 180°
which is straight line angle.
∴ AOB is a straight line.

Question 5.
In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = \(\frac {1}{2}\) (∠QOS – ∠POS)

Solution:
We have given,
Ray OR is perpendicular to line PQ
Therefore, ∠QOR = ∠POR = 90° ……(i)
Now, ∠QOS = ∠QOR + ∠ROS
or, ∠QOS = 90° + ∠ROS ……(ii) (from equation(i))
Again, ∠POS = ∠POR – ∠ROS
or, ∠PSO = 90 – ∠ROS ……(iii) (from equation (i))
Subtract equation (iii) from equation (ii),
∠QOS – ∠POS
∠QOS – ∠POS = 90° + ∠ROS – (90° – ∠ROS)
∠QOS – ∠POS = 90° + ∠ROS – 90° + ∠ROS
∠QOS – ∠POS = 2∠ROS
∠ROS = \(\frac {1}{2}\) (∠QOS – ∠POS)

Question 6.
It is given that ∠XYZ = 64° and xy is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Solution:
We have given that ∠XYZ = 64°
and ∠ZYQ = ∠QYP (∵ YQ is bisector of ∠ZYP)
Now, ∠XYZ + ∠ZYQ + ∠QYP = 180° (because XYP is a straight line)
or, 64° + ∠ZYQ + ∠ZYQ = 180° (∵ ∠ZYQ + ∠QYP)
or, 2∠ZYQ = 180° – 64° = 116°
or, ∠ZYQ = 58°
Now, ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58°
∠XYQ = 122°
Again, ∠QYP = ∠ZYQ (from equation (i))
∴ ∠QYP = 58° (∵ ∠ZYQ = 58°)
∴ Reflex ∠QYP = 360° – 58° = 302°